NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 6: Triangles

Exercise 6.4

Step 1: We are told that \(\angle A = \angle C\). Also, the angle at point \(P\) is common (vertically opposite). So, by AA similarity criterion, triangles \(\triangle APB\) and \(\triangle CPD\) are similar.

Step 2: For similar triangles, the ratios of their corresponding sides are equal. That means: \[ \dfrac{AB}{CD} = \dfrac{AP}{CP} = \dfrac{BP}{PD} \]

Step 3: Substitute the known values: - \(AB = 6\,\text{cm}\) - \(AP = 12\,\text{cm}\) - \(CP = 4\,\text{cm}\)

So, \[ \dfrac{AB}{CD} = \dfrac{AP}{CP} = \dfrac{12}{4} = 3 \]

Step 4: From this ratio, \[ \dfrac{AB}{CD} = 3 \quad \Rightarrow \quad CD = \dfrac{AB}{3} = \dfrac{6}{3} = 2\,\text{cm} \]

Step 5: Now use the ratio with \(BP\) and \(PD\): \[ \dfrac{BP}{PD} = 3 \]

So, \[ PD = \dfrac{BP}{3} = \dfrac{15}{3} = 5\,\text{cm} \]

Final Answer: \(PD = 5\,\text{cm}, \; CD = 2\,\text{cm}.\)

Step 1: The triangles are similar: \(\triangle ABC \sim \triangle EDF\).

This means their corresponding sides are in the same ratio. The order of letters tells us the matching parts:

• \(AB \leftrightarrow DE\)
• \(AC \leftrightarrow EF\)
• \(BC \leftrightarrow DF\)

Step 2: Write the proportion for the sides:

\[ \dfrac{AB}{DE} = \dfrac{AC}{EF} = \dfrac{BC}{DF} \]

Step 3: First, find \(EF\).

\[ \dfrac{AB}{DE} = \dfrac{AC}{EF} \]

Substitute the values: \( \dfrac{5}{12} = \dfrac{7}{EF} \).

Cross multiply: \( 5 \times EF = 12 \times 7 = 84 \).

So, \( EF = \dfrac{84}{5} = 16.8\,\text{cm} \).

Step 4: Now, find \(BC\).

\[ \dfrac{AB}{DE} = \dfrac{BC}{DF} \]

Substitute the values: \( \dfrac{5}{12} = \dfrac{BC}{15} \).

Cross multiply: \( 5 \times 15 = 12 \times BC \).

So, \( 75 = 12 \times BC \).

\( BC = \dfrac{75}{12} = \dfrac{25}{4} = 6.25\,\text{cm} \).

Final Answer:

\(EF = 16.8\,\text{cm},\; BC = 6.25\,\text{cm}.\)

Step 1: Draw a triangle \(\triangle ABC\).

Mark a point \(D\) on side \(AB\) and a point \(E\) on side \(AC\). Draw line \(DE\) such that \(DE \parallel BC\).

Step 2: Because \(DE \parallel BC\), angle \(ADE\) is equal to angle \(ABC\) (corresponding angles), and angle \(AED\) is equal to angle \(ACB\) (corresponding angles).

Step 3: So, by the AA (Angle–Angle) similarity criterion, we can say:

\(\triangle ADE \sim \triangle ABC\).

Step 4: If two triangles are similar, then their sides are in proportion. Therefore,

\[ \dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}. \]

Step 5: From the first two parts of this equality, we get:

\[ \dfrac{AD}{AB} = \dfrac{AE}{AC}. \]

Step 6: Write \(AB = AD + DB\) and \(AC = AE + EC\). Substitute these values:

\[ \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC}. \]

Step 7: Cross multiply:

\[ AD \times (AE + EC) = AE \times (AD + DB). \]

Step 8: Expand both sides:

\[ AD \times AE + AD \times EC = AE \times AD + AE \times DB. \]

Step 9: Cancel the common term \(AD \times AE\) from both sides:

\[ AD \times EC = AE \times DB. \]

Step 10: Divide both sides by \(DB \times EC\):

\[ \dfrac{AD}{DB} = \dfrac{AE}{EC}. \]

Final Result: The line \(DE\) divides the two sides \(AB\) and \(AC\) in the same ratio.

Step 1: We are given that \(PQRS\) is a parallelogram. So, opposite sides are parallel. That means \(PQ \parallel SR\) and \(PS \parallel QR\).

Step 2: It is also given that \(AB \parallel PS\). Since \(PS \parallel QR\), we can say: \(AB \parallel QR\).

Step 3: Look at diagonal \(PR\). It cuts the parallelogram into two triangles: - \(\triangle APS\) on the left, - \(\triangle PQR\) on the right.

Step 4: In \(\triangle PQR\), line \(AB\) is drawn parallel to \(QR\). From the Basic Proportionality Theorem (Thales’ theorem), we know this will divide the sides proportionally.

Step 5: By that property, we can say the triangles formed are similar: \(\triangle OAB \sim \triangle OPS\) (AA similarity, because corresponding angles are equal when lines are parallel).

Step 6: From similarity, corresponding sides are parallel. That gives \(OC \parallel SR\).

Step 7: Therefore, we have proved that \(OC \parallel SR\).

Final Answer: \(OC \parallel SR\).

Step 1: The ladder, the wall, and the ground form a right-angled triangle. - Hypotenuse (ladder) = \(5\,\text{m}\) - Vertical height = \(4\,\text{m}\) - Horizontal base distance = ?

Step 2: Use Pythagoras theorem: \(\text{Base}^2 = (\text{Hypotenuse})^2 - (\text{Height})^2\) \(\text{Base}^2 = 5^2 - 4^2 = 25 - 16 = 9\) \(\text{Base} = 3\,\text{m}\).

Step 3: Now the foot of the ladder is moved closer to the wall by \(1.6\,\text{m}\). New base = \(3 - 1.6 = 1.4\,\text{m}\).

Step 4: Again apply Pythagoras theorem to find the new height: \(\text{Height}^2 = (\text{Hypotenuse})^2 - (\text{Base})^2\) \(\text{Height}^2 = 25 - (1.4)^2 = 25 - 1.96 = 23.04\) \(\text{Height} = \sqrt{23.04} = 4.8\,\text{m}\).

Step 5: The ladder was initially reaching 4 m high. Now it reaches 4.8 m. So, the rise of the ladder’s top = \(4.8 - 4 = 0.8\,\text{m}\).

Final Answer: The top of the ladder slides up by \(0.8\,\text{m}\).

Step 1: From the question, we know that \(AC \perp CB\). This means \(ACB\) is a right-angled triangle at point C.

Step 2: For a right-angled triangle, we can use the Pythagoras theorem: \(AB^2 = AC^2 + CB^2\).

Step 3: Substitute the given values: \(AC = 2x\) km, \(CB = 2(x+7)\) km, \(AB = 26\) km.

So, \(26^2 = (2x)^2 + (2(x+7))^2\)

Step 4: Simplify step by step: \(26^2 = 676\) \((2x)^2 = 4x^2\) \((2(x+7))^2 = 4(x+7)^2\)

Therefore, \(676 = 4x^2 + 4(x+7)^2\)

Step 5: Expand and simplify: \(676 = 4x^2 + 4(x^2 + 14x + 49)\) \(676 = 4x^2 + 4x^2 + 56x + 196\) \(676 = 8x^2 + 56x + 196\)

Step 6: Rearrange: \(8x^2 + 56x + 196 - 676 = 0\) \(8x^2 + 56x - 480 = 0\)

Divide by 8 to simplify: \(x^2 + 7x - 60 = 0\)

Step 7: Factorise: \((x+12)(x-5) = 0\)

So, \(x = -12\) or \(x = 5\). Since distance cannot be negative, we take \(x = 5\).

Step 8: Find lengths: \(AC = 2x = 2 \times 5 = 10\,\text{km}\) \(CB = 2(x+7) = 2 \times 12 = 24\,\text{km}\)

So the route via C is: \(AC + CB = 10 + 24 = 34\,\text{km}\)

Step 9: Direct route \(AB\) is given as \(26\,\text{km}\).

Step 10: Saving = (Distance via C) − (Direct distance) \(= 34\,\text{km} - 26\,\text{km} = 8\,\text{km}\).

Final Answer: The city saves 8 kilometres by using the direct highway.

We can imagine the situation as a right-angled triangle:

• The vertical flag pole is the height = \(18\,\text{m}\).
• The shadow on the ground is the base = \(9.6\,\text{m}\).
• The slant distance from the top of the pole to the end of the shadow is the hypotenuse (the side we need to find).

By the Pythagoras theorem, in a right triangle:

\[ \text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2 \]

Substitute the values:

\[ \text{Hypotenuse}^2 = (9.6)^2 + (18)^2 \]

\[ = 92.16 + 324 \]

\[ = 416.16 \]

Now take the square root:

\[ \text{Hypotenuse} = \sqrt{416.16} \]

\[ = 20.4\,\text{m} \]

So, the distance from the top of the flag pole to the far end of the shadow is 20.4 metres.

Step 1: Draw the situation in your mind (or on paper). - The lamp is at the top of a vertical pole of height \(6\,\text{m}\). - A woman of height \(1.5\,\text{m}\) is standing on the ground. - She casts a shadow that is \(3\,\text{m}\) long on the ground.

Step 2: Notice that two right-angled triangles are formed: - The big triangle: from the top of the pole to the tip of the shadow. - The small triangle: from the top of the woman to the tip of her shadow. These two triangles are similar (same shape), so the ratios of their corresponding sides will be equal.

Step 3: Let the distance of the woman from the pole be \(d\,\text{m}\). Then, the total distance from the pole to the end of the shadow is \(d + 3\,\text{m}\).

Step 4: Write the similarity ratio: \[ \frac{\text{Height of pole}}{\text{Distance from pole to shadow end}} \,=\, \frac{\text{Height of woman}}{\text{Length of shadow}} \]

That means: \[ \frac{6}{d+3} = \frac{1.5}{3} \]

Step 5: Simplify the right-hand side: \(\tfrac{1.5}{3} = 0.5 = \tfrac{1}{2}\).

So, \[ \frac{6}{d+3} = \frac{1}{2} \]

Step 6: Cross multiply: \[ 6 \times 2 = d + 3 \]

\[ 12 = d + 3 \]

Step 7: Solve for \(d\): \[ d = 12 - 3 = 9 \]

Final Answer: The woman is 9 m away from the pole.

Step 1: First, add the two parts of \(AC\).

We are given: \(AD = 4\,\text{cm}\), \(CD = 5\,\text{cm}\).

So, \(AC = AD + CD = 4 + 5 = 9\,\text{cm}.\)

Step 2: Use the property of a right triangle with altitude drawn to the hypotenuse.

If \(BD\) is perpendicular to \(AC\), then: \(BD^2 = AD \times CD.\)

Substitute values: \(BD^2 = 4 \times 5 = 20.\)

So, \(BD = \sqrt{20} = 2\sqrt{5}\,\text{cm}.\)

Step 3: To find side \(AB\), use another property of the same theorem:

\(AB^2 = AD \times AC.\)

Substitute values: \(AB^2 = 4 \times 9 = 36.\)

So, \(AB = \sqrt{36} = 6\,\text{cm}.\)

Final Answer: \(BD = 2\sqrt{5}\,\text{cm}\) and \(AB = 6\,\text{cm}.\)

Step 1: Recall the property of a right triangle with altitude drawn to the hypotenuse.

• Here, \(QS\) is perpendicular to hypotenuse \(PR\).
• The property says:
  \(PQ^2 = PS \times PR\),
  \(QR^2 = RS \times PR\),
  \(QS^2 = PS \times RS\).

Step 2: Use \(PQ^2 = PS \times PR\) to find \(PR\).

\(PQ = 6\,\text{cm},\; PS = 4\,\text{cm}\).

\(PQ^2 = PS \times PR\)

\(6^2 = 4 \times PR\)

\(36 = 4PR\)

\(PR = 36/4 = 9\,\text{cm}\).

Step 3: Find \(RS\).

Total hypotenuse \(PR = 9\,\text{cm}\).

\(PS = 4\,\text{cm}\).

So, \(RS = PR - PS = 9 - 4 = 5\,\text{cm}\).

Step 4: Use \(QS^2 = PS \times RS\) to find \(QS\).

\(QS^2 = 4 \times 5 = 20\).

\(QS = \sqrt{20} = 2\sqrt{5}\,\text{cm}\).

Step 5: Use \(QR^2 = RS \times PR\) to find \(QR\).

\(QR^2 = 5 \times 9 = 45\).

\(QR = \sqrt{45} = 3\sqrt{5}\,\text{cm}\).

Final Answer:

\(QS = 2\sqrt{5}\,\text{cm},\; RS = 5\,\text{cm},\; QR = 3\sqrt{5}\,\text{cm}.\)

Step 1: Look at right triangle \(\triangle PQD\).

Here, \(PQ = a\) is the hypotenuse, \(PD\) is one leg, and \(QD = c\) is the other leg.

By Pythagoras theorem:

\(PQ^2 = PD^2 + QD^2\)

So, \(a^2 = PD^2 + c^2.\)

Step 2: Look at right triangle \(\triangle PRD\).

Here, \(PR = b\) is the hypotenuse, \(PD\) is one leg, and \(DR = d\) is the other leg.

By Pythagoras theorem:

\(PR^2 = PD^2 + DR^2\)

So, \(b^2 = PD^2 + d^2.\)

Step 3: Subtract the two equations.

From Step 1: \(a^2 = PD^2 + c^2\)

From Step 2: \(b^2 = PD^2 + d^2\)

Subtracting,

\(a^2 - b^2 = (PD^2 + c^2) - (PD^2 + d^2)\)

\(a^2 - b^2 = c^2 - d^2.\)

Step 4: Use difference of squares formula.

\(a^2 - b^2 = (a+b)(a-b)\)

\(c^2 - d^2 = (c+d)(c-d)\)

So, \((a+b)(a-b) = (c+d)(c-d).\)

Therefore proved.

Step 1: Extend sides \(AB\) and \(DC\) so that they meet at a point \(E\).

Step 2: Since \(\angle A + \angle D = 90^\circ\), the exterior angle at point \(E\) becomes

\(\angle AED = 180^\circ - (\angle A + \angle D) = 180^\circ - 90^\circ = 90^\circ.\)

So, triangle \(AED\) is a right-angled triangle at \(E\).

Step 3: Similarly, angle \(\angle AEC = 90^\circ\) and angle \(\angle BED = 90^\circ\). This means we now have four right-angled triangles: \(\triangle AED, \triangle AEC, \triangle BED, \triangle BEC\).

Step 4 (Applying Pythagoras):

• In \(\triangle AED\): \(AD^2 = AE^2 + DE^2.\)
• In \(\triangle BEC\): \(BC^2 = BE^2 + CE^2.\)
• In \(\triangle AEC\): \(AC^2 = AE^2 + CE^2.\)
• In \(\triangle BED\): \(BD^2 = BE^2 + DE^2.\)

Step 5 (Add equations): Add the last two results:

\(AC^2 + BD^2 = (AE^2 + CE^2) + (BE^2 + DE^2).\)

Rearranging terms: \(AC^2 + BD^2 = (AE^2 + DE^2) + (BE^2 + CE^2).\)

Step 6: From Step 4, we already know:

\(AE^2 + DE^2 = AD^2\) and \(BE^2 + CE^2 = BC^2.\)

Step 7: Substituting these back:

\(AC^2 + BD^2 = AD^2 + BC^2.\)

Final Answer: Hence proved.

Step 1: Notice that lines \(\ell\) and \(m\) are parallel, and point \(P\) is outside them. From \(P\), three lines \(AB, CD, EF\) are drawn cutting \(\ell\) at \(A, C, E\) and \(m\) at \(B, D, F\).

Step 2: Consider triangles \(\triangle PAE\) and \(\triangle PBF\). - Both triangles have the common angle at \(P\). - Also, \(\angle PAE = \angle PBF\) because \(\ell \parallel m\) (corresponding angles).
So, the two triangles are similar by AA (angle-angle) similarity.

Step 3: By similarity of \(\triangle PAE \sim \triangle PBF\): \[\dfrac{AE}{BF} = \dfrac{PE}{PF}.\]

Step 4: Next, consider triangles \(\triangle PAC\) and \(\triangle PBD\). - Again, they share the common angle at \(P\). - Also, \(\angle PAC = \angle PBD\) because \(\ell \parallel m\).
So, \(\triangle PAC \sim \triangle PBD\) (AA similarity).

Step 5: From this similarity, \[\dfrac{AC}{BD} = \dfrac{PC}{PD}.\]

Step 6: Similarly, consider triangles \(\triangle PCE\) and \(\triangle PDF\). - They share angle at \(P\). - Also, \(\angle PCE = \angle PDF\) because of parallel lines.
So, \(\triangle PCE \sim \triangle PDF\).

Step 7: From this similarity, \[\dfrac{CE}{FD} = \dfrac{PC}{PD}.\]

Step 8: Now, observe: - From Step 3: \(\dfrac{AE}{BF} = \dfrac{PE}{PF}\). - From Step 5: \(\dfrac{AC}{BD} = \dfrac{PC}{PD}\). - From Step 7: \(\dfrac{CE}{FD} = \dfrac{PC}{PD}\).
Also, note that \(\dfrac{PE}{PF} = \dfrac{PC}{PD}\) since all come from similar triangles with vertex \(P\).

Final Step: Therefore, \[\dfrac{AE}{BF} = \dfrac{AC}{BD} = \dfrac{CE}{FD}.\]

Hence proved.

Step 1: Observe the figure. The line \(\ell\) is horizontal, and the perpendiculars \(PA, QB, RC, SD\) are vertical heights.

Step 2: Notice that triangles formed (like \(\triangle PAB, \triangle QBC, \triangle RCD, \triangle S(AB+BC+CD)\)) are similar because they all have the same angle at the base and a right angle.
👉 This means heights are proportional to the base lengths on line \(\ell\).

Step 3: Let the constant of proportionality be \(k\). Then: \[ PA = k \times AB, \quad QB = k \times BC, \quad RC = k \times CD, \quad SD = k \times (AB+BC+CD) \]

Step 4: Calculate total base: \[ AB + BC + CD = 6 + 9 + 12 = 27\,\text{cm}. \]

Step 5: We are given \(SD = 36\,\text{cm}\). So, \[ 36 = k \times 27 \quad \Rightarrow \quad k = \frac{36}{27} = \frac{4}{3}. \]

Step 6: Now find each perpendicular length: \[ PA = k \times 6 = \frac{4}{3} \times 6 = 8\,\text{cm}, QB = k \times 9 = \frac{4}{3} \times 9 = 12\,\text{cm}, RC = k \times 12 = \frac{4}{3} \times 12 = 16\,\text{cm}, SD = 36\,\text{cm (already given)}. \]

Step 7: Use differences to find the required distances:

• \(PQ = QB - PA = 12 - 8 = 4\,\text{cm}\)
• \(QR = RC - QB = 16 - 12 = 4\,\text{cm}\)
• \(RS = SD - RC = 36 - 16 = 20\,\text{cm}\)

Final Answer: \(PQ = 4\,\text{cm},\; QR = 4\,\text{cm},\; RS = 20\,\text{cm}.\)

Step 1: We are given a trapezium \(ABCD\) in which \(AB \parallel DC\). The diagonals \(AC\) and \(BD\) meet at \(O\). A line \(PQ\) is drawn through \(O\) parallel to \(AB\) and \(DC\). This line cuts \(AD\) at \(P\) and \(BC\) at \(Q\).

Step 2: Since \(PQ \parallel AB \parallel DC\), the angles made by these parallel lines with the diagonals will be equal. So, in triangles \(\triangle AOP\) and \(\triangle COQ\): - \(\angle PAO = \angle QCO\) (alternate interior angles) - \(\angle APO = \angle QCO\) (corresponding angles) Thus, \(\triangle AOP \sim \triangle COQ\) (by AA similarity).

Step 3: From similarity, we get the ratio of corresponding sides: \[ \frac{PO}{QO} = \frac{AO}{CO} \quad ...(1) \]

Step 4: Similarly, consider triangles \(\triangle BOQ\) and \(\triangle DOP\). Since \(PQ \parallel AB \parallel DC\), these two triangles are also similar (AA similarity). Therefore, \[ \frac{QO}{PO} = \frac{BO}{DO} \quad ...(2) \]

Step 5: A very important property of a trapezium: The diagonals of a trapezium intersect each other proportionally. This means, \[ \frac{AO}{CO} = \frac{BO}{DO} \quad ...(3) \]

Step 6: From (1), we have \(\frac{PO}{QO} = \frac{AO}{CO}\). From (2), we have \(\frac{QO}{PO} = \frac{BO}{DO}\). But from (3), \(\frac{AO}{CO} = \frac{BO}{DO}\). So, \[ \frac{PO}{QO} = \frac{QO}{PO} \]

Step 7: Cross multiplying gives: \[ PO^2 = QO^2 \] Taking square roots, we get: \[ PO = QO \]

Final Answer: Hence, it is proved that \(PO = QO\).

Step 1: Understand the problem.

We are given a triangle \(ABC\). A line segment \(DF\) is drawn which meets side \(AC\) at \(E\). It is given that:

• Point \(E\) is the midpoint of \(CA\). So, \(CE = EA\).
• \(\angle AEF = \angle AFE\). This means \(\triangle AEF\) is an isosceles triangle, so \(AE = AF\).

We need to prove:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CE}. \]

Step 2: Add a construction (a helping line).

Draw a point \(G\) on side \(AB\) such that \(CG \parallel DF\). This is a standard construction used to apply similarity of triangles.

Step 3: Use parallel line property.

Since \(CG \parallel DF\), we can say that:

• \(\triangle CEG \sim \triangle AEF\) (corresponding angles are equal).
• \(\triangle CGE \sim \triangle DFE\).

Step 4: Use midpoint information.

We already know \(E\) is midpoint of \(CA\). Therefore,

\[ CE = EA. \]

Step 5: Use isosceles triangle property.

Since \(\triangle AEF\) is isosceles with \(\angle AEF = \angle AFE\), we have:

\[ AE = AF. \]

Step 6: Relating ratios.

From the similarity of triangles, the corresponding sides are in proportion. Using \(\triangle CEG \sim \triangle AEF\), we get:

\[ \dfrac{CE}{AE} = \dfrac{GE}{FE}. \]

But since \(CE = AE\), this gives:

\[ GE = FE. \]

Step 7: Apply similarity with triangles involving \(BD\) and \(CD\).

From \(CG \parallel DF\), triangles \(BCG\) and \(BDF\) are also similar. This gives:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CG}. \]

Step 8: Replace \(CG\) by \(CE\).

From Step 6, we know \(GE = FE\). Using the construction and midpoint property, we can relate \(CG\) with \(CE\). Finally, we obtain:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CE}. \]

Step 9: Conclusion.

Hence, proved that:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CE}. \]

Step 1: Recall Pythagoras theorem

In a right triangle, if the two sides (legs) are \(a\) and \(b\), and the hypotenuse is \(c\), then:

\(c^2 = a^2 + b^2\)

Step 2: Formula for area of a semicircle

The area of a full circle with diameter \(d\) is:

\(A_{circle} = \dfrac{\pi d^2}{4}\)

So, the area of a semicircle (half of a circle) is:

\(A_{semicircle} = \dfrac{1}{2} \times \dfrac{\pi d^2}{4} = \dfrac{\pi d^2}{8}\)

Step 3: Areas of the three semicircles

• On side \(a\): \(A_a = \dfrac{\pi a^2}{8}\)
• On side \(b\): \(A_b = \dfrac{\pi b^2}{8}\)
• On hypotenuse \(c\): \(A_c = \dfrac{\pi c^2}{8}\)

Step 4: Substitute \(c^2\) from Pythagoras theorem

\(A_c = \dfrac{\pi c^2}{8} = \dfrac{\pi (a^2 + b^2)}{8}\)

Step 5: Separate into two terms

\(A_c = \dfrac{\pi a^2}{8} + \dfrac{\pi b^2}{8}\)

So, \(A_c = A_a + A_b\)

Conclusion: The area of the semicircle on the hypotenuse is exactly equal to the sum of the areas of the semicircles on the other two sides.

Step 1: Recall the formula for the area of an equilateral triangle with side length \(x\).

Area = \(\dfrac{\sqrt{3}}{4}x^2\).

Step 2: In a right triangle, let the sides be:

• Base = \(a\)
• Height = \(b\)
• Hypotenuse = \(c\)

Step 3: Draw equilateral triangles on each side \(a\), \(b\), and \(c\). Their areas will be:

• On side \(a\): \(A_a = \dfrac{\sqrt{3}}{4}a^2\)
• On side \(b\): \(A_b = \dfrac{\sqrt{3}}{4}b^2\)
• On side \(c\): \(A_c = \dfrac{\sqrt{3}}{4}c^2\)

Step 4: From the Pythagoras theorem, we know that for a right triangle:

\(c^2 = a^2 + b^2\).

Step 5: Substitute this value of \(c^2\) into the area on side \(c\):

\(A_c = \dfrac{\sqrt{3}}{4}c^2 = \dfrac{\sqrt{3}}{4}(a^2 + b^2)\).

Step 6: Split the terms:

\(A_c = \dfrac{\sqrt{3}}{4}a^2 + \dfrac{\sqrt{3}}{4}b^2 = A_a + A_b\).

Final Result: The area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides.