Question

please solve question from 1 to 5

Answer 1

[The pic is of 1st no.]
1.Then 
area = 1/2× b×h
864=30h
h= 271.566
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2.GIVEN: ABC an isosceles triangle, AB=AC. So angle B = angle C. Base BC = 24cm,

TO FIND: The perimeter(AB + BC + AC) =?

CONSTRUCTION: AM perpendicular to BC.

Since triangle AMB is congruent to triangle AMC( by RHS congruence criterion)

So, M is the mid point of BC. So, BM = 12cm

Since area(triangle ABC) = 1/2 * BC * AM

=> ar(tri ABC) = 1/2 * 24 * AM = 192

=> AM = 192/ 12 = 16cm.

Now in right triangle AMB

AB² = AM² + BM²

=> AB² = 16² + 12²

=> AB² = 256 + 144 = 400

=> AB = √400 = 20cm

So, AC = 20cm

So, perimeter = 20+20+24 = 64cm ……….ANS
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3.Consider the triangle ABC
AB=34
BC=20
CA=42
=>S=(34+20+42)/2
       =48
remember that area of a triangle=√s(s-a)(s-b)(s-c)
                                             =√48*14*28*6
                                              =336cm²
area of parallelogram=2*336=672cm²
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4.Picture is attached.
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5.Let the two legs of the triangle be

x

and

y

.

Then

x+y+5=12

and

x2

+

y2

=

52

y=7−x

Substituting:

x2

+

(7−x)

2

=25

x2

+49−14x+

x2

=25

2

x2

−14x+24=0

2

(

x2

−7x+12)

=0

2

(x−3)

(x−4)

=0

x=3and4

The two legs measure

3

and

4

cm. The area is given by

A=

B×H

2

, where

B

and

H

are the two legs.

A=

3×4

2

A=6

square centimetres.

In summary:

The triangle has sides of

3,4and5

centimetres. The area is of

12

cm2

Hopefully this helps!

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