A Parallelogram and a Triangle between the Same Parallels

Q.

A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangles and parallelogram is

1. 1:2

2. 2:1

3. 1:1

4. 1:3

Q. Question 2
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that: $ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)$.

Q. Question 4 (i)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
$\left(i\right)ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)$

Q.

If a triangle and a square are on the same base and between the same parallels, then the ratio of area of triangle to the area of square is

1. 3 : 1

2. 1 : 3

3. 1 : 4

4. 1 : 2

Q. ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm² and ar (ΔABOC) = 6 cm², then area of rectangle ABCD is

(a) 9 cm²

(b) 12 cm²

(c) 15 cm²

(d) 18 cm²

Q.

If a triangle and a parallelogram are on same base and between same parallels then the ratio of the areas of the triangle to the area of parallelogram is:

1. 1 : 2

2. 1 : 3

3. 1 : 4

4. 3 : 1

Q.

For two figures to be on the same base and between the same parallels , they must have a common base and.

1. One common vertex

2. Two common vertices

3. The vertices(or the vertex) opposite to the common base lying on a line making an acute angle to the base

4. The vertices(or the vertex) opposite to the common base lying on a line parallel to the base

Q.

IN the figure, PQRS is a parallelogram. If X and Y are mid points of PQ and SR respectively and diagonal SQ is joined, the ratio $ar\left(||gmXQRY\right):ar\left(△QSR\right)$=

1. 1: 4

2. 2:1

3. 1: 2

4. 1:1

Q.

In a triangle and a parallelogram are on the same base and between the same parallels then the ratio of the area of the triangle to the area of the parallelogram is

(a) 1:2

(b) 1:3

(c) 1:4

(d) 3:4

Q. Question 3
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).

Q. Question 1 (iii)
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA ( see the given figure). AC is a diagonal. Show that:

PQRS is a parallelogram

Q.

Paralallelogram $ABCD$, triangle $ABE$ and rectangle $ABFG$ are on the same base $AB$ and between same parallel lines. Area of parallelogram is ${}^{\prime }{p}^{\prime }$, area of triangle is ${}^{\prime }{t}^{\prime }$ and area of rectangle is ${}^{\prime }{r}^{\prime }$. Find the relation between $p,t$ and $r$.

1. $p=r=\frac{t}{2}$

2. $p=r=t$

3. $p=r=2t$

4. $p=\frac{R}{2}=\frac{t}{2}$

Q. Question 2
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that: $ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)$.

Q. Question 8
D, E and F are the mid – points of the sides BC, CA and AB, respectively of an equilateral $\Delta ABC$, show that $\Delta DEF$ is also an equilateral triangle.

Q. Question 4 (ii)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that:
(ii) ar(APD) + ar(PBC) = ar(APB) +ar(PCD)

Q. Question 8
In trapezium ABCD, AB || DC and L is the midpoint of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. prove that ar (ABCD) = ar (APQD).

Q. If ABCD is a parallelogram, then prove that

ar (ΔABD) = ar (ΔBCD) = ar (ΔABC) = ar (ΔACD) = $\frac{1}{2}$ ar (||^gm ABCD)

Q.

For two figures to be on the same base and between the same parallels, they must have a common base and

1. One common vertex

2. The vertices(or the vertex) opposite to the common base lying on a line making an acute angle to the base

3. The vertices(or the vertex) opposite to the common base lying on a line parallel to the base

4. Two common vertices

Q.

In the given figure, ABCD is a parallelogram. E and F are midpoints of BC and AD respectively. The ratio of area $\Delta AFE$ : area of parallelogram ABCD is

1. 1 : 4

2. 1 : 8

3. 4 : 1

4. 8 : 1

Q.

In the given figure area of $\Delta DAC$ is $60c{m}^2$, then the area of of parallelogram ABCD is

1. $240c{m}^2$
2. $120c{m}^2$
3. $80c{m}^2$
4. $40c{m}^2$

Q. Question 12 (iii)
ABCD is a trapezium in which AB | | CD and AD =BC (see the given figure). Show that $\Delta ABC\cong \Delta BAD$.

Q. Question 4 (ii)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that:
(ii) ar(APD) + ar(PBC) = ar(APB) +ar(PCD)

Q. Question 4 (i)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
$\left(i\right)ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)$

Q. Question 12 (iv)
ABCD is a trapezium in which AB || CD and AD =BC (see the given figure). Show that Diagonal AC=diagonal BD.

Q.

In the figure if DE = BF = 7 cm and AC = 9 cm. Find the area of the parallelogram.

1. $63c{m}^2$

2. $126c{m}^2$

3. $31.5c{m}^2$

4. $70c{m}^2$

Q.

In the adjoining figure, ABCD and BQSC are two parallelograms.Prove that $ar\left(△RSC\right)=ar\left(△PQB\right)$.

Q. In the given figure, ABCD is a parallelogram, if M is any point that lies on AD when produced, then which of the following is not true?

1. 
   $ar\left(\Delta ABD\right)=ar\left(\Delta ACB\right)$
2. 
   $ar\left(\Delta ABC\right)=\frac{1}{2}ar\left(ABCD\right)$
   
3. 
   $ar\left(\Delta AMC\right)=2ar\left(\Delta ADC\right)$
4. 
   $ar\left(MBD\right)=ar\left(\Delta DCM\right)$

Q. If a triangle and parallelogram are on the same base and between the same parallels, then the ratio of area of the triangle to that of area of the parallelogram is ___.

Q.

In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 6 cm, AE = 8cm and CF = 12 cm, find AD.
[2 MARKS]

Q. Prove that area of a $\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{altitude}$.