हिंदी

In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = 12 ar (∆ABC). - Mathematics

Advertisements
Advertisements

प्रश्न

In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = `1/2` ar (∆ABC).

योग

उत्तर


Given: In ∆ABC, D is the mid-point of AB and P is any point on BC.

CQ || PD means AB in Q.

To prove: ar (∆BPQ) = `1/2` ar (∆ABC)

Construction: Join PQ and CD.

Proof: Since, D is the mid-point of AB. So, CD is the median of ∆ABC.

We know that, a median of a triangle divides it into two triangles of equal areas.

∴ ar (∆BCD) = `1/2` ar (∆ABC)

⇒ ar (∆BPD) + ar (∆DPC) = `1/2` ar (∆ABC)  ...(i)

Now, ∆DPQ and ∆DPC are on the same base DP and between the same parallel lines DP and CQ.

So, ar (∆DPQ) = ar (∆DPC)  ...(ii)

On putting the value from equation (ii) in equation (i), we get

ar (∆BPD) + ar (∆DPQ) = `1/2` ar (∆ABC)

⇒ ar (∆BPQ) = `1/2` ar (∆ABC)

Hence proved.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.3 [पृष्ठ ९०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.3 | Q 4. | पृष्ठ ९०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)


In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.


In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]


P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

(i) ar(PRQ) = 1/2 ar(ARC)

(ii) ar(RQC) = 3/8 ar(ABC)

(iii) ar(PBQ) = ar(ARC)


In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :

(1) ar (Δ PBQ) = ar (Δ ARC)

(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)

(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .


ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.


X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)


The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)


A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)


In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×