Question

25
4. ABCD is a rectangle. Taking AD as a dia-
meter a semicircle AXD is drawn which
intersects diagonal BD at X. If AB= 12 cm,
AD = 9 cm, find the values of (i) BD (ii) BX.​

Answer 1

Answer:

BD = 15cm & BX = 48/5

Step-by-step explanation:

(1)

In ABD,

∠ABD = 90              (as ABCD is a rectangle)

By Pythagoras theorem,

AD² + AB² = BD²

=>  9²+12²=BD²

=>  81 + 144 = BD²

=>  225 = BD²

=>  √225 = BD

=>  15 cm = BD

(2)

Ar.(ΔABD) = Ar.(ΔABD)

=>  1/2 × BD × AX = 1/2 × AD × AB      (as in a circle, angle subtended by the diameter is always equal to 90)

=>  BD × AX = AD × AB

=>  15 × AX = 9 × 12

=>  AX = 108/15

=>  AX = 36/5      ------------------(1)

As, ∠AXD is 90, so will be ∠AXB

In ΔAXB,

by Pythagoras Theorem,

AX² + BX² = AB²

=>  (36/5)² + BX² = 12²

=>  BX² = 12²- (36/5)²

=>  BX² = 144- (1296/25)

=>  BX² = (3600 - 1296)/25

=>  BX² = 2304/25

=>  BX = √(2304/25)

=>  BX = 48/5 cm

I HOPE IT HELPS!

Answer 2

BD=15

BX =48/5

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