NCERT Mathematics Class 9 Exemplar Ch 5 Introduction to Euclid's Geometry Part 7

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∠ABC = ∠ ACB, ∠ 3 = ∠ 4. Show that ∠ 1 = ∠ 2

∠ABC = ∠ ACB, ∠ 3 = ∠ 4. Show That ∠ 1 = ∠ 2

∠ABC = ∠ ACB, ∠ 3 = ∠ 4. Show that ∠ 1 = ∠ 2

9. In the Fig, we have , . Show that .

Answer: Given that: and According to Euclid’s axioms, if equal are subtract from equals, then reminders are also equal. On subtracting equation (ii) from equation (i), we get

AC = DC, CB = CE. Show that AB = DE

AC = DC, CB = CE. Show That AB = DE

AC = DC, CB = CE. Show that AB = DE

10. In the Fig. , we have , . Show that .

Answer: Given that: and According to Euclid’s axioms, if equals are added to equals, the then wholes are also equal. So, on adding equation (i) and equation (ii), we get

ABC = ? ACB, ? 3 = ? 4. Show that ? 1 = ? 2

ABC = ? ACB, ? 3 = ? 4. Show That ? 1 = ? 2

ABC = ? ACB, ? 3 = ? 4. Show that ? 1 = ? 2

11. In the Fig, if , and , show that .

Answer: Given that: and and According to Euclid’s axioms, things which are double of the same things are equal to one another. On multiplying equation (iii) by 2, we get [From (i) and (ii)]

12. In the Fig.:

(i) , M is the mid-point of AB and N is the mid- point of BC. Show that .

(ii) , M is the mid-point of AB and N is the mid-point of BC. Show that .

AB = BC, M mid-point of ABand N mid- point of BC

AB = BC, M Mid-Point of ABand N Mid- Point of BC

AB = BC, M mid-point of ABand N mid- point of BC

Answer: (i) Given that: is the mid – point of AB. ∴ and N is the mid – point of BC ∴ According to Euclid’s axioms, things which are halves of the same things are equal to one another. From Equation (i), we get On multiplying both sides by , we get [using (ii) and (iii)] (ii) Given that: M is the mid –point of AB ∴ and N is the mid – point of BC ∴ According to Euclid’s axioms, things which are doubles of the same things are equal to one another. On multiplying both sides of equation (i) by 2, we get [using (ii) and (iii)]

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