Question

Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm
respectively. The length of the side of largest square FDGB that can be inscribed in the
triangle ABE is

(a) 32/3cm
(b) 16/3cm
(c)8/3cm
(d) 4/3cm
BOARD SAMPLE PAPER QUESTION​

Answer 1

Answer:

Hope it helps ❤

option b is the correct option..

Answer 2

Given:

The right-angled at $B$ is of length $16cm$ and $8cm$.

To find:

The length of the side of the largest square $FDGB$inscribed in the triangle$ABE$ is

Step-by-step explanation:

for parallel lines $FD$ and $CE$

with transversal $AE$

∠$ADF=$∠$AEB$

Also, $FDGB$ is a square

∠$AFD=90^{0}$, ∠$DGF=90^{0}$

Δ$AFD$=Δ$DGF$

Δ$AFD$≈Δ$DGF$

The sides in the similar triangle are proportional

$\frac{AF}{DG}=\frac{FD}{GE}$

$\frac{16-x}{x} =\frac{x}{8-x}$

$(16-x)(8-x)=x^{2}$

$16(8-x)-x(8-x)=x^{2}$

$128-16x-8x+x^{2}=x^{2}$

$128-16x-8x=0$

$128-24x=0$

$128=24x$

$x=\frac{128}{24}$

$x=\frac{16}{3}cm$

Answer:

Therefore, The largest square$FDGB$ that can be inscribed in the

triangle $ABE$ is $\frac{16}{3}cm$.