geometry homework

Question

1) A triangle has side lengths of 6, 9, and 11. Decide whether it is an acute, right, or obtuse triangle. Explain

2) If EFGH is a triangle, EF=7 and EH=4, what value must FH have?

3) Find the altitude of an isosceles triangle whose base is 10 and whose congruent sides are 9.

4) What is the length of the diagonal  of a square whose side lengths are 7/2?

5) What is the lengthof an altitude of an equilateral triangle whose side length is 8/3?

6) Rectangle PQRS with QR=8 and SQ=10 (SQ cuts thru the middle), What is the value of y or PQ?

Answer 1

1)Given that 6 , 9 , 11 are the lengths of the sides of a triangle.

Here a = 6 and b = 9 and c = 11.

Hence the triangle is obtuse.

3) That bisecting altitude line will split the base into two equal parts, 5 & 5.

From there we can exploit your Pythagorean theorem .

.
'a' & 'b' are your sides, and 'c' is your hypotenuse .

(Note Pythagorean theorem will ONLY work on right triangles, or triangles with an angle that is exactly 90˚).

Given that congruent side is 9.

Altitude = 7.48

Answer 2

6)Rectangle PQRS, SQ cuts the middle.it divides the rectangle as two equal right angled triangles.

The equal triangles are SRQ,PQS.

From pythogorus thm

SQ^2 =PQ^2+SP^2

10^2 = PQ^2+8^2

100 = PQ^2+64

100-64 = PQ^2

36 = PQ^2

⇒PQ = 6

 

Answer 3

5)Altitude of an equilater triangle =a√3/2

=(8/3)(√3/2)

=4/√3

Answer 4

2).

Let, EFH is a right triangle.

From pythagorean theorem :

EF^2 = EH^2 + FH^2

Substitute 4 for EH and 7 for EF.

7^2 = 4^2 + FH^2

49 = 16 + FH^2

FH^2 = 49 - 16 = 33

FH = sqrt(33) = 5.744.

Therefore, FH = 5.744.

Answer 5

4).

Length of the side of the square(a) = 7/2.

Diagonal of the square(d) = a√2, where a = Length of the side of the square.

d = (7/2) * √2 = 7/√2.

Therefore, diagonal of a square is 7/√2.