The vertex is given at (2,-1), which tells us that h= and k=
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The area of the trapezoidal comes out to be 125 square units.
<h3>What is trapezoidal?</h3>A four-sided shape called a trapezoid has one pair of parallel sides. It fundamentally resembles a square, rectangle, or parallelogram in two dimensions.
Now, according to the question;
If a trapezoid PQRS of parallel sides PQ & RS has always been divided into 4 triangles by it's own diagonals PR and QS, which intersect at X,
Then, the area for triangle PSX equals which of triangle QRX, as well as the product of triangle PSX but also triangle QRX equals the same of triangle PQX & triangle RSX.
Let area of the triangle PSX is A₁
Let area of the triangle QRX is A₂
Let area of the triangle PQX is A₃
Let area of the triangle RSX is A₄
Then,
A₁ × A₂ = A₃ × A₄
As A₁ = A₂
The,
A₁² = A₃ × A₄
The areas for the triangles;
A₃ = 20 square units
A₄ = 45 square units.
Substitute the values;
A₁² = 20 × 45
= 900
A₁ = 30
Thus, A₁ = A₂ = 30 square units.
The entire area of the trapezium is;
= A₁ + A₂ + A₃ + A₄
= 30 + 30 + 20 + 45
= 125
Therefore, the complete area of the trapezium PQRS is 125 square units.
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The correct question is-
In trapezoid PQRS, PQ is parallel to RS. Let X be the intersection of diagonals PR and QS. The area of triangle PQX is 20 and the area of triangle RSX is 45. Find the area of trapezoid PQRS.
