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liubo4ka [24]
3 years ago
15

Do Now:

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

Area = 32

Step-by-step explanation:

Remark

The standard area for a Triangle is

Area = 1/2 * b * h

Givens

b = 16

k = 1/2

h = 4

Solution

Area = 1/2 * b * h

Area = 1/2 * 16 * 4

Area = 32

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In parallelogram QRST QS RT Is QRST a rectangle ?
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Answer:

The correct option is A.

Step-by-step explanation:

It is given that the QRST is a parallelogram and the diagonals QS and RT are congruent.

According to the property of parallelogram if the length of diagonals are same, then the parallelogram is rectangle. We can also prove it.

Let QRST be an parallelogram and QS=RT.

In triangles QRT and RQS.

RT=QS         (Diagonals)

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So by SSS rule of congruence, triangle QRT and triangle RQS are congruent.

By CPCT,

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\angle R=90

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Similarly in triangle QST and RTS,

RT=QS         (Diagonals)

ST=ST         (Common side)

QT=RS         (Opposite sides of the parallelogram are same)

So by SSS rule of congruence, triangle QST and RTS are congruent.

By CPCT,

\angle S=\angle T

Since the sum of two consecutive angles of a parallelogram is 180 degree.

Therefore the measure of angle S and angle T are 90 degree.

Since the measure of all interior angles of QRST are 90 degree, therefore the parallelogram QRST  is a rectangle.

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