Question

ADVANCED LEVEL where n and 6. The figure shows a triangle STU where TU = 11 cm. H lies on TU such that the length of TH is 120% of the length of HU and ZSHU = 90°. Given that the area of ASTH is 21 cm?, find ZTSU. S Irigo the d bodi prod T U H -11 cm​

Answer 1

Step-by-step explanation:

The area of a triangle having sides a,b,c and S as semi-perimeter is given by

A=S(S−a)(S−b)(S−c)

S=2a+b+c=211+60+61=7132=66cm

We need to find the altitude to smallest side

A=66(66−11)(66−60)(66−61)

⇒A=66,55,6.5

⇒A=106900

⇒A=330m2

The area of a triangle having base AC and height P/S given by

Area (A) = 21=21 (Base × Height)

Area (A)= 21(AC×P)

We need to find the height P. Here the smaller side is 11 m.

AC= 11 m

330=21.(11×p)

330×2=11×p

660=11×p

∴p=660/11

p=60 m.

Answer 2

Answer:

310

Step-by-step explanation:

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