A school has triangular shaped ground with vertices P(0 , 4) ,Q(-5 , 0) and R(5 , 0).The type of triangular ground is * 1 point Equilateral Scalene Isosceles Right-angled triangle
Answers
Answer:
C
Step-by-step explanation:
Correct option is
C
Isosceles triangle
Let the points A(-4,0),B(4,0)and C(0,3)are the vertices
∴AB=
(4−(−4))
2
+(0−0)
2
=
(8)
2
=
64
=8
∴BC=
(0−4)
2
+(0−3)
2
=
16+9
=
25
=5
∴CA=
(−4−0)
2
+(0−3)
2
=
16+9
=
25
=5
As we know that two points are equal in the given vertices
∴This is the vertices of a isosceles triangle.
Given : A school has triangular shaped ground with vertices P(0 , 4) , Q(-5 , 0) and R(5 , 0).
To Find : The type of triangular ground is
Equilateral
Scalene
Isosceles
Right-angled triangle
Solution:
P(0 , 4) , Q(-5 , 0) and R(5 , 0)
PQ = √(-5 - 0)² + ( 0 - 4)² = √25 + 16 = √41
PR = √( 5 - 0)² + ( 0 - 4)² = √25 + 16 = √41
RQ = √(-5 - 5)² + ( 0 - 0)² = √100 + 0 = 10
√41 = √41 ≠ 10
Two sides of triangle are equal
Hence Triangle is isosceles
(√41 )² + (√41)² = 82 ≠ (10)² Hence not a right angle triangle
So Correct option is Isosceles
Equilateral Triangle : All 3 sides Equal
Isosceles Triangle : Any 2 sides are equal
Scalene Triangle : No sides are equal
Right angle triangle : Sum of Square of two smaller side = Square of Larger side ( or one angle is right angle 90°)
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