Question

In Figure-3, PQ || BC, PQ = 3 cm, BC = 9 cm and AC = 7-5 cm. Find the
length of AQ.​

Answer 1

Answer:

attatch the photo of the figure so that we can answer it

Answer 2

Q:Find the area of the quadrilateral ABCD whose vertices are A(-4,-3),

Find the area of the quadrilateral ABCD whose vertices are A(-4,-3),B(3,-1), C(0,5) and D(-4,2).

GIVEN:

A(-4,-3),B(3,-1), C(0,5) and D(-4,2).are the vertices Quadrilateral.

NOW:

The quardrilateral turned out to be into 2 traiangles by joining diagonal BD OR AC

ABD.......traiangle 1

BDC.......2

WE KNOW THAT

$area \: of \: triangle = \frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|$

TRIANGLE ABD

$\frac{1}{2} | - 4( - 1 - 2) + 3(2 - ( -3) + ( - 4)( - 3 - ( - 1)| \\ = \frac{1}{2} | - 4( - 3) + 3(2 + 3) - 4( - 3 + 1)| \\ = \frac{1}{2} |12 + 3(5) -4( - 2) | \\ = \frac{1}{2} |12 + 15 + 8| \\ = \frac{1}{2} |35 \\ = \frac{35}{2} |$

TRIANGLE BDC

$\frac{1}{2} |3(2 - 5) + ( - 4)(5 - ( - 1) + 0( - 1 - 2)| \\ = \frac{1}{2} |3( - 3) - 4(5 + 1) + 0( - 3)| \\ \frac{1}{2} | - 9 - 4(6) + 0 | \\ \frac{1}{2} | - 9 - 24| \\ \frac{1}{2} | - 33| \\ \frac{33}{2}$

AREA OF THE QUADRILATERAL = ABD +BDC

• 35/2+33/2

=68/2

=34unit squares