Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then, construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.

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Solution

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.

The required triangle can be drawn as follows.

Step 1

Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.

Step 2

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.

Step 3

Draw a ray AX making an acute angle with AB, opposite to vertex C.

Step 4

Locate 5 points (as 5 is greater in 5 and 3), $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ , on line segment AX such that $A A_{1} = A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5}$

Step 5

Join $A_{3} B$ . Draw a line through $A_{5}$ parallel to $A_{3} B$ intersecting extended line segment AB at B'.

Step 6

Through B', draw a line parallel to BC intersecting extended line segment AC at C'. ΔAB'C' is the required triangle.

Justification

The construction can be justified by proving that

$A B^{'}$ = $\frac{5}{3}$ AB , $B^{'} C^{'}$ = $\frac{5}{3}$ BC , $A C^{'}$ = $\frac{5}{3}$ AC ,

In ΔABC and ΔAB'C',

∠ABC = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

∴ ΔABC ∼ ΔAB'C' (AA similarity criterion)

$\frac{A B}{A B^{'}}$ = $\frac{B C}{B^{'} C^{'}}$ = $\frac{A C}{A C^{'}}$ … (1)

In Δ $A A_{3} B a n d Δ A A_{5} B^{'}$ ,

∠ $A_{3} A B = ∠ A_{5} A B^{'}$ (Common)

∠ $A A_{3} B = ∠ A A_{5} B^{'}$ (Corresponding angles)

∴ Δ $A A_{3} B \sim Δ A A_{5} B^{'}$ (AA similarity criterion)

$\frac{A B}{A B^{'}}$ = $\frac{A A_{3}}{A A_{5}}$

$\frac{A B}{A B^{'}}$ = $\frac{3}{5}$ ....(9)

On comparing equations (1) and (2), we obtain

$\frac{A B}{A B^{'}}$ = $\frac{B C}{B^{'} C^{'}}$ = $\frac{A C}{A C^{'}}$ = $\frac{3}{5}$

$\frac{A B^{'}}{A B}$ = $\frac{5}{3}$ , $\frac{B^{'} C^{'}}{B C}$ = $\frac{5}{3}$ , $\frac{A C^{'}}{A C}$ = $\frac{5}{3}$

This justifies the construction.

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Similar questions

Q. Draw a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then, construct another triangle whose sides are

\frac{5}{3}

times the corresponding sides of the given triangle. [CBSE 2011]

Q. Draw a right triangle in which the sides (other than the hypotenuse) are of lengths 4 cm and 3 cm. Now construct another triangle whose sides are

\frac{3}{5}

times the corresponding sides of the given triangle.

Question 7
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths $4 c m$ and $3 c m$ . Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are times the corresponding sides of the given triangle. Give the justification of the construction.

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Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then, construct another triangle whose sides are 53 times the corresponding sides of the given triangle.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then, construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.