हिंदी

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is ______. - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • 1 : 3

  • 1 : 2

  • 3 : 1

  • 1 : 4

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is 1 : 2.

Explanation:

We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

i.e., Area of triangle = `1/2` Area of parallelogram

⇒ `"Area of triangle"/"Area of parallelogram " = 1/2`

∴ Area of triangle : Area of parallelogram = 1 : 2

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.1 [पृष्ठ ८७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.1 | Q 9. | पृष्ठ ८७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).


D and E are points on sides AB and AC respectively of ΔABC such that

ar (DBC) = ar (EBC). Prove that DE || BC.


The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that

ar (ABCD) = ar (PBQR).

[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]


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, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).


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.]


PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.


ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).


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).


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×