Question
The base of a triangle is 5√45 cm and the height is 3√88 cm. What is the area of the triangle? Recall: A= 1/2 bh
Responses
A 8√133 cm^2
B 12√87 cm^2
C 45√110 cm^2
D 75√22 cm^2
Responses
A 8√133 cm^2
B 12√87 cm^2
C 45√110 cm^2
D 75√22 cm^2
Answers
GPT-4o mini
To find the area of the triangle, we can use the formula:
\[
A = \frac{1}{2} \times b \times h
\]
where \( b \) is the base and \( h \) is the height.
Given:
- Base \( b = 5\sqrt{45} \) cm
- Height \( h = 3\sqrt{88} \) cm
Now we will first simplify \( b \) and \( h \):
1. Simplifying \( b = 5\sqrt{45} \):
\[
\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5}
\]
Therefore,
\[
b = 5 \cdot 3\sqrt{5} = 15\sqrt{5} \text{ cm}
\]
2. Simplifying \( h = 3\sqrt{88} \):
\[
\sqrt{88} = \sqrt{4 \times 22} = \sqrt{4} \cdot \sqrt{22} = 2\sqrt{22}
\]
Therefore,
\[
h = 3 \cdot 2\sqrt{22} = 6\sqrt{22} \text{ cm}
\]
Now substituting \( b \) and \( h \) into the area formula:
\[
A = \frac{1}{2} \times (15\sqrt{5}) \times (6\sqrt{22})
\]
Calculating the product:
\[
= \frac{1}{2} \times 15 \times 6 \times \sqrt{5} \times \sqrt{22} = \frac{1}{2} \times 90 \times \sqrt{110}
\]
\[
= 45 \sqrt{110} \text{ cm}^2
\]
Now, comparing with the provided options:
- A: \( 8\sqrt{133} \) cm²
- B: \( 12\sqrt{87} \) cm²
- C: \( 45\sqrt{110} \) cm²
- D: \( 75\sqrt{22} \) cm²
The correct area of the triangle is:
\[
\boxed{45\sqrt{110} \text{ cm}^2}
\] (Option C).
\[
A = \frac{1}{2} \times b \times h
\]
where \( b \) is the base and \( h \) is the height.
Given:
- Base \( b = 5\sqrt{45} \) cm
- Height \( h = 3\sqrt{88} \) cm
Now we will first simplify \( b \) and \( h \):
1. Simplifying \( b = 5\sqrt{45} \):
\[
\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5}
\]
Therefore,
\[
b = 5 \cdot 3\sqrt{5} = 15\sqrt{5} \text{ cm}
\]
2. Simplifying \( h = 3\sqrt{88} \):
\[
\sqrt{88} = \sqrt{4 \times 22} = \sqrt{4} \cdot \sqrt{22} = 2\sqrt{22}
\]
Therefore,
\[
h = 3 \cdot 2\sqrt{22} = 6\sqrt{22} \text{ cm}
\]
Now substituting \( b \) and \( h \) into the area formula:
\[
A = \frac{1}{2} \times (15\sqrt{5}) \times (6\sqrt{22})
\]
Calculating the product:
\[
= \frac{1}{2} \times 15 \times 6 \times \sqrt{5} \times \sqrt{22} = \frac{1}{2} \times 90 \times \sqrt{110}
\]
\[
= 45 \sqrt{110} \text{ cm}^2
\]
Now, comparing with the provided options:
- A: \( 8\sqrt{133} \) cm²
- B: \( 12\sqrt{87} \) cm²
- C: \( 45\sqrt{110} \) cm²
- D: \( 75\sqrt{22} \) cm²
The correct area of the triangle is:
\[
\boxed{45\sqrt{110} \text{ cm}^2}
\] (Option C).