Question

please answer my question as fast as possible

Answer 1

33.


In Triangle AXQ and triangle BXQ
∠XQA = ∠XMB (both 90°)
QX =QX (common)
AM = BM (as it's being bisected)

Therefore, By SAS congurence criteria Triangle AXQ is congruent to triangle BXQ

AX = XB (by CPCT)

Now in triangle XAY and triangle XBY

XA = XB(poved above)
YX =YX(common)
∠AXY = ∠YXB (both equal to 90°+40° by external amgle of triangel)

Hence, Triangle XAY is congruent to Triangle XBY

Therefore,
∠XAY = ∠XBY
∠XAY = 10°

Now in triangle
AYQ
∠YMA + ∠AYQ + ∠YAX + ∠XAQ = 180° ( angle sum property)

=> 90° +∠ AYX + 10° + 40° = 180°
=> ∠AYX = 180°- 140°
Hence, ∠AYX = 40°.



34.

Let those equal sides be x
Now by using Pythagoras Theorem

(AB)^2 + ( BC)^2 = (AC)^2
x^2 + x^2 =128
=> 2x^2 = 128
=> x^2 = 64

Hence x = 8 units.

Therefore AB = 8 units.



35.

As all three triangles are congruent.
So,
∠ACB =∠ACH = ∠BCH

Therefore,
∠ACB + ∠ACH + ∠BCH = 360°(angles around a point)
3 ∠BCH = 360°
Hence, ∠BCH = 120°.

Answer 2

Answer:

Q number 34 ka answer - 8

Q number 35 ka answer - 120 degree