Question

plse help me to solve this sum

Answer 1

the two triangles are right triangles, use Pythagorean theorem

Answer 2

they are asking you to find the area, not the sum.....
find the area for each triangle
find the area for the rectangle in the middle and add the three

Answer 3

do something similar for the 2nd part

create a parallelogram and 1 triangle

Answer 4

make sense?

Answer 5

yes , let me solve again
thanks

Answer 6

opps the first one is screwed up....
those right triangles cant have the same base if they have different hypotenuse

their bases together has to be 15, but the longer hypotenuse has to have a longer base

Answer 7

use this for the 1st one

the height for each triangle should be the same

Answer 8

why not 7.5

Answer 9

the lines drawn down are perpendicular to the bottom base.
if one hypotenuse is 14 and the other hypotenuse is 13 and the heights are the same
the base of each triangle must be different

Answer 10

if you use 7.5 you will get a different height for each triangle, which is not possible if that is a rectangle in the middle.
Trapezoid the bases have to be parallel

Answer 11

what would be the height of traingles

Answer 12

c^2 - a^2 = b^2
14^2-8.4^2 = 11.2

13^2 -6.6^2 = 11.2

Answer 13

does that make sense

Answer 14

I get area of 196

Answer 15

in the 2nd one your triangle will have a height of 11.2, and a base of 15

parallelgram will have width of 10 and length of 14

Answer 16

\[area of trapezoid=\frac{ 25+10 }{ 2 }h=17.5h\]
also area=10*h+\[\frac{ 1 }{ 2 }*15*h\]
\[area=\frac{ 20+15 }{2 }h=17.5h\]
area in both cases is same.

Answer 17

\[h ^{2}=13^{2}-x ^{2},\]
\[also h ^{2}=14^{2}-\left( 15-x \right)^{2}\]
\[169-x ^{2}=196-\left( x ^{2} -30x+225\right)\]
\[169-x ^{2}=196-x ^{2}+30x-225\]
169-196+225=30x
30x=198
\[x=\frac{ 198 }{ 30 }=6.6\]
now you can find the h and area