Download App

Areas Of Parallelograms And Triangles — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsAreas Of Parallelograms And Triangles5 Marks
Question
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
  1. Prove that $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{ECF}).$
  2. If the area of $\triangle\text{DFB}=3\text{cm}^2,$ find the area of $||^{gm}ABCD$.
✓

Answer

In triangles ADF and ECF, we have$\angle\text{ADF}=\angle\text{ECF}$ [Alternate interior angles, Since AD || BE]
AD = EC [since AD = BC = CE] And $\angle\text{DFA}=\angle\text{CFA}$ [Vertically opposite angles] So, by AAS congruence criterion, we have$\triangle\text{ADF}\cong\triangle\text{ECF}$
$\Rightarrow\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{ECF})$ and DF = CF.
Now, DF = CF ⇒ BF is a median in $\triangle\text{BCD}.$
$\Rightarrow\text{ar}(\triangle\text{BCD})=2\text{ar}(\triangle\text{BDF})$
$\Rightarrow\text{ar}(\triangle\text{BCD})=2\times3\text{cm}^2=6\text{cm}^2$
Hence, area of a parallelogram $=2\text{ar}(\triangle\text{BCD})=2\times6\text{cm}^2=12\text{cm}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Mark Question" from Areas Of Parallelograms And Triangles→Areas Of Parallelograms And Triangles — all question types→Maths STD 9 question papers→Maharashtra Board STD 9 question papers→Maharashtra Board question paper generator→

Similar questions

Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder : $(2m^2 – 3m + 10) ÷ (m – 5)$
→
In the given figure, compute the value of x.
→
In $\triangle\text{ABC,}$ side AB is produced to D such that BD = BC. If $\angle\text{B}=60^{\circ},$ and $\angle\text{B}=60^{\circ},$ prove that:
  1. AD > CD and
  2. AD > AC.
→
In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,Image
i. $\angle DEG =\frac{1}{2} \angle EDF$
ii. $EF = FG$
→
In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD
→
In the figure, $AB\ ||\ PQ.$ Find the value of $x$ and $y$.

→
Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ∠QPR respectively. Ray PB and ray PA are perpendicular to each other.
Draw a figure showing all these rays and write –
i. A pair of complementary angles
ii. A pair of supplementary angles
iii. A pair of congruent angles.
→
The price of 3 chairs and 2 tables is ₹ 4500 and price of 5 chairs and 3 tables is ₹ 7000, then find the price of 2 chairs and 2 tables.
→
The sum of the deviations of a set of n values $x_1, x_2, x_3,..., x_n$ measured from $15$ and $-3$ are $-90$ and $54$ respectively. Find the value of n and mean.
→
The following data gives the value (in crores of rupees) of the Indian export of cotton textiles for different years:
Years
1982–83
1983–84
1984–85
1985–86
1986–87
Value of exports of cotton textiles (in crores of rupees)
300
325
475
450
550
Represent the above data with the help of a bar graph. Indicate with the help of a bar graph the year in which the rate of increase in exports is maximum over the preceding year.
→