Question

Please help by solving with method. Thanks ​

Answer 1

Let H be the point where the height starting from B meets the base DC.

BHC is a right triangle, and we have

$HC=DH-AB=15-7=8$

Using the Pythagorean theorem, we have

$BH = \sqrt{BC^2-HC^2} = \sqrt{100-64} = 6$

Finally, since BH=AD, we have

$AD=6$

The area is given by the usual formula

$A = \dfrac{(B+b)h}{2} = \dfrac{6(15+7)}{2} = 3(15+7)=3\cdot 22 = 66$

And the perimeter is simply

$2p = AB+BC+CD+AD = 7+10+15+6 = 38$

Answer 2

Answer:

(a) AD=6 cm

(b) 66 cm²

(c) 38 cm

Step-by-step explanation:

(a) To get the length of AD, we can use pythagoras on the triangle part. Let's say the dotted line intersects DC at E. Then since AB and DC are parallel, AD is equal to BE. EC = 8 cm simply by subtracting AB from DC (15-7=8).

Then pythagoras says that 8²+BC²=10². So BC = √(100-64) = 6. And since BC=AD, AD is also 6.

(b) Area of the rectangle DEBA is 7*6 = 42. Area of triangle ECB is base times half height: 8*3 = 24. Sum is 42+24 = 66 cm²

(c) Perimeter is 7+10+15+6 = 38 cm. Add all the side lengths.