Question

11. Find the area of the shaded region in Fig. 17.12
12 cm
52 cm
16 cm
48 cm
Fig. 17.12
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Answer 1

We know that Δ ADB is a right angled triangle.

According to the Pythagoras theorem, the area of the square on the hypotenuse is equal to the sum of the areas of the squares of its remaining two sides.

Formula: AD² + DB² = AB² {substituting the values}

12² + 16² = AB²

144 + 256 = AB²

400 = AB²

√400 = AB

AB = 20 cm

Area of ΔADB = √s(s-a)(s-b)(s-c)

$s$ = $\frac{12+16+20}{2}cm$

$s$ = $\frac{48}{2}cm$

$s$ = $24cm$

$a$ = $12cm$

$b$ = $16cm$

$c$ = $20cm$

√s(s-a)(s-b)(s-c)

√24(24-12)(24-16)(24-20) cm²

√24(12)(8)(4) cm²

√9216 cm²

96 cm²

Area of ΔABD = 96 cm²

Area of ΔABC = √s(s-a)(s-b)(s-c)

$s$ = $\frac{52+48+20}{2}cm$

$s$ = $\frac{120}{2}cm$

$s$ = $60cm$

$a$ = $52cm$

$b$ = $48cm$

$c$ = $20cm$

√s(s-a)(s-b)(s-c)

√60(60-52)(60-48)(60-20) cm²

√60(8)(12)(40) cm²

√230400 cm²

480 cm²

Area of ΔABC = 480 cm²

Area of the shaded region (quadrilateral ACBD) = Area of ΔABC - Area of ΔABD

Area of quadrilateral ACBD = 480 cm² - 96 cm²

Area of quadrilateral ACBD = 384 cm²

Therefore, the area of the shaded region is 384 cm².

Answer 2

Step-by-step explanation:

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